Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 327: 16

$y'= cos^2(x^4)4x^3$

We are told to find the derivative using Part 1 of the Fundamental Theorem of Calculus given the expression: $y = \int ^{x^4}_{0}cos^2(θ)dθ$ Since the upper bound is a function we need to use the chain rule: $u = x^4$ $u' = 4x^3$ $y = \int^u_0 cos^2(u)du$ Since $F(u) = \int^{u}_{0}f(t)dt$ $F'(u) = f(u)$ Thus, $y' = cos^2(u)u'$ $y'= cos^2(x^4)4x^3$